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No, one cannot obtain a solution of this PDE by adding a solution of heat equation to a solution of the wave equation. (With linear PDE, we can combine solutions of the same equation to make new ones; but your situation is different). Your PDE is known as damped wave equation and is solved here. It does inherit some features from the heat and wave equation. ...

3

As you pointed out we have \begin{align*} -\Delta w & = 0 \text{ in }\Omega \\ w & = 0 \text{ on }\partial\Omega \end{align*} and \begin{align*} -\Delta (-w) & = 0\text{ in }\Omega \\ -w & = 0\text{ on }\partial\Omega \end{align*} Thus we can use the maximum principle for $w$ and $-w$, yielding $\max_\Omega w = \max_{\partial\Omega} w =0$ and ...

2

Here are three (somewhat standard) references: Jurgen Jost, "Riemannian Geometry and Geometric Analysis." Peter Li, "Lecture Notes on Geometric Analysis." Thierry Aubin, "Nonlinear Analysis on Manifolds. Monge-Ampere Equations." Jost's book is on its sixth edition. Aubin's book has a first and second edition, although my understanding is that the first ...

2

You are correct to use separation of variables, the next step is to solve the two independent differential equations. First, I believe that you should use $\frac {X''}{X} = \frac{4T'}{T} = - \lambda^2$ as this gives you a negative only answer. This will give you the following, $X(x) = A\sin( \lambda x)+B\cos( \lambda x)$ $T(t) = Ce^{\frac {-\lambda^2 ... 2 I'm guessing you mean$u\in C^2(\Omega)\cap C^0(\bar\Omega)$.$u$is a continuous function on$\bar\Omega$, which is compact, hence reaches its extrema. Let$x_0$be a maximum point. If$x_0\in\Omega$, then$Du(x_0)=0$and$\Delta u(x_0)\leq 0$. By plugging this into your equation, we get $$\underbrace{\Delta u(x_0)}_{\leq 0} + ... 2$$ u(x,t) = V(x)\mathrm{e}^{st +ikx} $$and$$ u_t =u_{xx} + \mu u $$we find$$ su = \left(V''\mathrm{e}^{st +ikx} +2ikV'\mathrm{e}^{st +ikx}-k^2u\right) + \mu u $$thus$$ \left(V'' +2ikV'\right)\mathrm{e}^{st +ikx}+(\mu -k^2-s)u = 0 $$This is what I would of done without the answer you have shown, make the terms in the bracket zero.$$ \mu -k^2-s = 0 ... 2 Hint:$\left(\dfrac{\partial u}{\partial x}\right)^2-\left(\dfrac{\partial u}{\partial y}\right)^2+\dfrac{1}{2}\dfrac{(y^2-x^2)^2}{xy(x^2+y^2)}\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial y}=02xy(x^2+y^2)\left(\dfrac{\partial u}{\partial x}\right)^2+(x^2-y^2)^2\dfrac{\partial u}{\partial x}\dfrac{\partial u}{\partial ...

2

It depends... the Laplacian is always defined with respect to some metric: $\Delta f = \nabla \cdot \nabla f$ and divergence requires a metric. Alternatively, you can define $\Delta$ as the gradient, in the sense of the calculus of variations, of the Dirichlet energy $\int \langle \nabla f, \nabla f\rangle\,dV$ and here again the metric is seen. My guess is ...

2

The change $$u(t,x)=e^{ax+bt}\,v(t,x)$$ for appropriate choice of $a,b\in\mathbb{R}$ will transform the equation into $$v_t=v_{xx},\quad v(0,x)=e^{-ax}\,g(x).$$

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The full procedure is to set both sides equal to an arbitrary constant $\lambda$. Then we have to solve $f''(x) = \lambda f(x)$ with the boundary conditions $f(0)=f(1)=0$, and there are three cases: $\lambda > 0$ gives the hyperbolic case, and (as you can check for yourself) the only function in this family of solutions which satisfies the boundary ...

2

Just plugin the definition of $T$ and $Q_\sigma$. We have $$u = \sigma Tu \iff Tu = \frac{u}{\sigma}$$ that is, by definition of $T$ iff $v := \frac{u}\sigma$ is the (unique) solution of $$a^{ij}(x,u,Du)D_{ij}v + b(x,u,Du) = 0, \quad v|_{\partial \Omega} = \phi$$ Let here $v = \frac u\sigma$, this gives $$\frac 1\sigma a^{ij}(x,u,Du) D_{ij}u + ... 2 Answer for part (a) should be$$\frac{1-ik}{1+k^2}$$So by the inversion formula$$f(x)=\frac{1}{2\pi}\int^{\infty}_{-\infty}\frac{1-ik}{1+k^2} e^{ikx}\,dkNow separate the integrand into real part and imaginary part. Notice that f(x) is a real function, so the imaginary part in the integrand should be zero. The real part is exactly the integral in ... 1 Because (\Delta f,f) \le -\lambda_1\|f\|^{2} for f\in\mathcal{D}(-\Delta), then \begin{align} \frac{d}{dt}\|T(t)f\|^{2}& =(\Delta T(t)f,T(t)f)+(T(t)f,\Delta T(t)f) \\ & \le -2\lambda_1(T(t)f,T(t)f)= -2\lambda_1\|T(t)f\|^{2},\;\; f \in\mathcal{D}(-\Delta) \end{align} Hence, \frac{d}{dt}\left( e^{2\lambda_1 ...

1

Let's put the tutorial solution into a normalized form. Which means either using only hyperbolic functions or expressing everything exponential or hyperbolic in $e^x$ and $e^{-x}$. Aiming for hyperbolic functions, one gets, using $e^x=\cosh(x)+\sinh(x)$, \begin{align} u_e(x)&=\frac{3e+1}{\cosh(1)}\sinh(x)-3e^x+x-x^2-2 \\ \\ ...

1

First: The definition of conformality looks odd; did you mean something like $Dg(x)^{t} \cdot Dg(x) = C(x) I$, i.e., $$\left\langle Dg(x)u, Dg(x)v \right\rangle = C(x) \langle u, v\rangle \quad\text{for all u, v}$$ instead? Second: Inside the parentheses, note that $x^{t}x = \|x\|^{2}$.

1

Solve $$v_t-v_{xx}=0,\quad v(0,t)=v(\ell,t)=0,\quad v(x,0)=f(x)$$ and $$w_t-w_{xx}=g(x,t),\quad w(0,t)=w(\ell,t)=0,\quad w(x,0)=0.$$ The $u=v+w$. Standard separation of variables gives $v$ in the form $$v(x,t)=\sum_{n=1}^\infty a_n\,e^{-\bigl(\tfrac{k\,\pi}{\ell}\bigr)^2\,t}\,\sin\frac{k\,\pi\,x}{\ell}.$$ To find $w$ develop $g$ in a Fourier series ...

1

It looks like you have a slight mistake, in that $$\max_{\Omega}(-(u_1 - u_2)) = - \min_{\Omega}(u_1 - u_2)$$ not $+\min_{\Omega}(u_1 - u_2)$ which you have written. This means that $$\min_{\Omega}(u_1 - u_2) = -\max_{\partial\Omega}(-(g_1 - g_2)) = \min_{\partial \Omega}(g_1 - g_2).$$ You already ...

1

a more general solution to $u_x + u_y = 1$ is $u(x,y)= Ax + (1-A)y+C$ the use of the variable $y$ in the expression of the boundary condition is unfortunate , we could say $u(p, \frac{p}{2}) = p$ for all real numbers $p$. plug that into the general solution, keeping in mind that this is an identity in $p$, by equating coefficients you should get that $C=0$ ...

1

The Poisson kernel for balls with centre $0$ in $\mathbb{R}^n$ is given by $$P(x,\zeta) = \frac{1}{\lVert\zeta\rVert\omega_{n-1}} \cdot \frac{\lVert\zeta\rVert^2 - \lVert x\rVert^2}{\lVert \zeta - x\rVert^n},$$ and for any function $u$ harmonic on the open ball with radius $R$ and continuous on the closed ball, in particular for entire harmonic functions, ...

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For $H^1$- functions, this does not hold in general. It is basically the same argument that $L^2$- functions generally do not vanish at $\infty$, you just have to take such a function and integrate it, for example a bump function where the bumps get thinner when you go outside. For $H^2$-functions however, it does hold: Take $$\int_0^a ... 1 Use a sequence of harmonics with increasing frequency and decreasing amplitude:$$f_n(x)= A_n \sin \left(\frac{\pi n}{L} x\right)$$As long as A_n\to 0, the values at time T tend to zero in the L^p sense. On the other hand, the solution at time t<T is$$f_n(x)= A_n e^{\lambda (\pi n/L)^2 (T-t)}\sin \left(\frac{\pi n}{L} x\right)$$which, for ... 1 For every x\in \overline{\Omega'}, we have$$\sup \{ u(z) : z \in B_r(x)\} \leqslant 3^n\inf \{ u(z) : z \in B_r(x)\}\tag{1}$$by assumption. From the open cover \{ B_r(x) : x \in \overline{\Omega'}\} of \Omega' we extract a finite subcover \{ B_m : 1 \leqslant m \leqslant k\}. If B_m \cap B_j \neq \varnothing, then we have$$\sup \{ u(z) : z ...

1

$$F: \big(u(t),\lambda\big) \to \frac{d^2}{dt^2}u + \lambda \sin u$$ Expand $\sin u$ into series around $u=0$, get linearization of $F$ operator: $$F\big(u,\lambda\big) = \frac{d^2}{dt^2}u + \lambda \left( u - \frac{u^3}{3!} + \cdots\right) \approx \bigg[ \frac{d^2}{dt^2} + \lambda I\bigg] u,$$ where $I$ is an identity mapping. Denote $L:X\times R\to ... 1 Note that, with the fact that$a \neq b$, $$\frac{\log(b)}{\log(\frac{b}{a})} - \frac{\log(a)}{\log(\frac{b}{a})} = \frac{\log(b) - \log(a)}{\log(b) - \log(a)} = 1$$ This tells that $$\frac{n\pi\log(b)}{\log(\frac{b}{a})} = n\pi +\frac{n\pi\log(a)}{\log(\frac{b}{a})}$$ and that $$\cos(\frac{n\pi\log(b)}{\log(\frac{b}{a})}) = ... 1 No, the Laplacian does not map L^2 to L^2. But one can consider it as an unbounded operator from a dense subspace of L^2 into L^2. This dense subset includes C_0^\infty, since the Laplacian maps C_0^\infty into C_0^\infty. To discuss self-adjointness, one must impose boundary conditions and close the operator. I suggest reading a book about ... 1 You need to additionally assume that S(t) is a symmetric matrix for all t, and that it has even dimension 2k \times 2k. Here are some hints (without giving away the whole solution). Let’s first simplify notation (to avoid using the t variable in two different ways, which is confusing to me). Define:$$H(u, z) = (1/2)z^T S(u) z$$Thus: ... 1 Here's a summary of, and extending remarks and explanations to, the comments. Note that my description of u(t,x) is based on experimental observations, not formal proofs. I plotted the sum for u(t,x) for a few values of t: within a moderate range for x, it would converge, although I did need extended accuracy (used Maple for this). My understanding ... 1$$ |y-x|^2=|y|^2-2\,y\cdot x+|x|^2=1-2\,y\cdot x+|x|^2.  |x|^2\,\Bigl|\,y-\frac{x}{|x|^2}\Bigr|^2=|x|^2\Bigl(|y|^2-\frac{2\,y\cdot x}{|x|^2}+\frac{1}{|x|^2}\Bigr)=|x|^2-2\,y\cdot x+1.$\$

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